# Planck's derivation of Blackbody Radiation

I have been studying quantum mechanics(blackbody radiation basically) and came up with an equation in Planck's derivation(where he had to assume that the oscillators in the cavity walls were limited to energies of $$\epsilon_n = nh\nu,\:$$ where $$n = 0, 1, 2, 3, ...$$) that the average energy per standing wave in the cavity is given by $$\bar\epsilon = \frac{h\nu}{e^{h\nu/kT}-1}$$ instead of energy equipartition average of $$k_BT$$ which Rayleigh and Jeans had used by classical mechanics(Here, $$k_B$$ is the Boltzmann's constant).

How can we prove this equation of average energy $$\:\bar\epsilon\:$$ that Planck came up with? How Planck got this relation? The only thing I know is that from Maxwell-Boltzmann's distribution law, the number of oscillators with energy $$\:\epsilon_n\:$$ is proportional to $$exp[-\epsilon_n/k_BT]$$.

Thanks in advance for any help

The only thing I know is that from Maxwell-Boltzmann's distribution law, the number of oscillators with energy $$\epsilon_n$$ is proportional to $$\exp[−\epsilon_n/k_BT]$$.

More formally you can write the probability $$p_n$$ of the oscillator having energy $$\epsilon_n=nh\nu$$ as $$p_n=A\ e^{−nh\nu/k_BT} \tag{1}$$ where $$A$$ is a still unknown constant.

The second thing you know is that all probabilities $$p_n$$ must add up to $$1$$: $$\sum_{n=0}^\infty p_n = 1$$ Using the expression from (1) you get: $$\sum_{n=0}^\infty A\ e^{−nh\nu/k_BT} = 1 \tag{2}$$

From (2) you can calculate $$A$$. I omit the actual calculation and tell only the result: $$A=1-e^{-h\nu/k_BT}$$

Finally you have everything needed to calculate the average energy $$\bar{\epsilon}$$: \begin{align} \bar{\epsilon} &= \sum_{n=0}^\infty p_n \epsilon_n \\ &= \ ... \\ &= \frac{h\nu}{e^{h\nu/k_BT}-1} \end{align}

Again I have left the calculational details to you.

• Thank you very much, it helped me a lot May 22, 2021 at 21:04